course Mth 163 ????a???f????assignment #003 003. Precalculus I 06-19-2008
......!!!!!!!!...................................
08:45:48 `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:45:50 For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
08:45:53 `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:45:55 Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
08:46:00 `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:46:03 The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
08:46:06 `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:46:09 In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024).
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
08:56:37 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
......!!!!!!!!...................................
RESPONSE --> If we move one unit to right and one unit to the left we get x = 6.8182 and x = 4.8182. y = -0.45833*(6.8182^2)+5.33333*6.8182-6.875 y = 8.1818 y = -0.45833*(4.8182^2)+5.33333*4.8182-6.875 y = 8.1818 The y values differ from the y value at the vertex by 0.4584. This difference corresponds to the a value in the function. confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:58:02 06-19-2008 08:58:02 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
......!!!!!!!!...................................
NOTES -------> Remember that if we move one unit to the right or left of the vertex of a parabola we can find the difference in the y value by the coefficient a of x^2. We could use that as a check to make sure we've found the correct values for y.
.......................................................!!!!!!!!...................................
08:58:03 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:08:37 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
......!!!!!!!!...................................
RESPONSE --> For the model y = -1x^2 + 10x + 100 we find the vertex to be x = -10 / 2*-1 , x = 5 y = -1*(5^2) + 10(5) + 100, y = 125 The vertex is (5, 125) The same would apply here as it did in the last problem to finding the point one unit to the right and left of the vertex. y=-1*(4^2) + 10(4) + 100 = 124 Y = -1*(6^2) + 10(6) + 100 = 124 The coordinates would be (4, 124) and (6, 124). The graph would be a parabola with its highest point at (5, 125). It would open downward towards the x axis, descending slowly at first and then faster. It would cross the x axis. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:09:45 06-19-2008 09:09:45 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
......!!!!!!!!...................................
NOTES -------> Thats true that because a = -1, the graph is descending and that if it was positive the graph would be ascending.
.......................................................!!!!!!!!...................................
09:09:46 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
?|????E??????[~?Q? assignment #004 004. Precalculus I 06-19-2008
......!!!!!!!!...................................
09:20:20 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
......!!!!!!!!...................................
RESPONSE --> f(3) = 3^2 + 4 f(3) = 13 f(7) = 7^2 + 4 f(7) = 53 f(-5) = -5^2 + 4 f(-5) = 29 The coordinates for the graph would be (3, 13), (7, 53), and (-5, 29). The graph would be a parabola passing through these points. The vertex would be: -b / 2a = 0 / 2*1 = 0 f(0) = 0 ^2 + 4 = 4. The vertex of the graph would be (0, 4). The points one unit to the left or right of the vertex would be (1, 5) and (1, -5) and would continue on to the points found earlier. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:20:37 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:30:49 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
......!!!!!!!!...................................
RESPONSE --> We just substitute for each value: f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 = (x+2)(x+2) + 4 = x(x+2)+2(x+2) + 4 = x^2 + 4x + 8 f(x+h) = (x+h)^2 + 4 = (x+h)(x+h) + 4 = x^2 + 2xh + h^2 + 4 f(x+h) - f(x) = (x+h)^2 + 4 - f(x) = x^2 + 2xh + h^2 + 4 - f(x) [f(x+h) - f(x)] / h = [x^2 + 2xh + h^2 + 4 - f(x)] / h = x^2 / h + 2x + h + 4/h - f(x)/h confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:35:29 06-19-2008 09:35:29 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
......!!!!!!!!...................................
NOTES -------> For the last two expression I did not put the expression for f(x) in to solve. Therefore, it looks like I just did a couple steps and stopped before solving the entire function. Need to remember when it has an expression like f(x+h) - f(x) that I have to substitute for both f(x) values into the expression.
.......................................................!!!!!!!!...................................
09:35:30 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:45:13 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
......!!!!!!!!...................................
RESPONSE --> Now we just substitute each expression into f(x) = 5x + 7: f(x1) = 5x1 + 7 f(x2) = 5x2 + 7 [f(x2) - f(x1)] / (x2 - x1) = [ 5x2 + 7 - ( 5x1 + 7)] / (x2 - x1) = ( 5x2 - 5x1 ) / (x2 - x1) = We can factor out 5 and get 5 ( x2 - x1) / (x2 - x1) = 5 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:46:00 06-19-2008 09:46:00 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
......!!!!!!!!...................................
NOTES -------> I still need to practice the expressions like the third that has all the parts to it.
.......................................................!!!!!!!!...................................
09:46:01 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:49:02 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
......!!!!!!!!...................................
RESPONSE --> If we want to find what value of x is f(x) = -3 we would write the expression like this: 5x + 7 = -3 We would just solve for this function to get the value: 5x + 7 - 7 = -3 - 7 5x = -10 x = -2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:49:31 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
`questionNumber 30000 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
......!!!!!!!!...................................
RESPONSE --> If we move one unit to right and one unit to the left we get x = 6.8182 and x = 4.8182. y = -0.45833*(6.8182^2)+5.33333*6.8182-6.875 y = 8.1818 y = -0.45833*(4.8182^2)+5.33333*4.8182-6.875 y = 8.1818 The y values differ from the y value at the vertex by 0.4584. This difference corresponds to the a value in the function. confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:58:02 06-19-2008 08:58:02 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
......!!!!!!!!...................................
NOTES -------> Remember that if we move one unit to the right or left of the vertex of a parabola we can find the difference in the y value by the coefficient a of x^2. We could use that as a check to make sure we've found the correct values for y.
.................................................
......!!!!!!!!...................................
08:58:03 `questionNumber 30000 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:08:37 `questionNumber 30000 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
......!!!!!!!!...................................
RESPONSE --> For the model y = -1x^2 + 10x + 100 we find the vertex to be x = -10 / 2*-1 , x = 5 y = -1*(5^2) + 10(5) + 100, y = 125 The vertex is (5, 125) The same would apply here as it did in the last problem to finding the point one unit to the right and left of the vertex. y=-1*(4^2) + 10(4) + 100 = 124 Y = -1*(6^2) + 10(6) + 100 = 124 The coordinates would be (4, 124) and (6, 124). The graph would be a parabola with its highest point at (5, 125). It would open downward towards the x axis, descending slowly at first and then faster. It would cross the x axis. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:09:45 06-19-2008 09:09:45 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
......!!!!!!!!...................................
NOTES -------> Thats true that because a = -1, the graph is descending and that if it was positive the graph would be ascending.
.................................................
......!!!!!!!!...................................
09:09:46 `questionNumber 30000 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
......!!!!!!!!...................................
RESPONSE -->